Question

1. Доведи, що бісектриси кутів паралелограма, які прилягають до однієї зі сторін, перетинаються під кутом 90°. пожалуйста срочно! даю 40 балов​

Answer 1

Відповідь:

Пояснення:

Дано:

АВСD - паралелограм;

АМ - бісектриса ∠А;

ВК - бісектриса ∠В.

Довести: АМ⊥ ВК

Доведення:

У паралелограма сума кутів, прилеглих до однієї сторони дорівнює 180*:

∠А + ∠В = 180*

Бісектриса кута ділить кут навпіл.

Точка О - точка перетину бісектрис АМ і ВК.

Розглянемо трикутник АОВ:

∠ВАО = 1/2•∠А

∠АВО = 1/2•∠В

∠ВАО + ∠АВО = 1/2•∠А + 1/2•∠В =

= 1/2•(∠А + ∠В) = 1/2•180 = 90*;

Сума кутів трикутника становить 180*.

∠АОВ = 180 - (∠ВАО + ∠АВО) = 180-

- 90= 90*
АМ⊥ ВК.

Отже. бісектриси кутів паралелограма, які прилягають до однієї зі сторін, перетинаються під кутом 90°.
Доведено.

Answer 2

Bisectors of Angles in a Parallelogram

In a parallelogram, the bisectors of the angles that are adjacent to one side intersect at a right angle (90°). Let's prove this statement.

Consider a parallelogram ABCD, where AB is parallel to CD and AD is parallel to BC. Let's label the points where the bisectors of angles A and B intersect side AD as E and F, respectively.

To prove that the bisectors of angles A and B intersect at a right angle, we need to show that angle EDF is 90°.

Proof:

1. Since AB is parallel to CD, angle A and angle C are alternate interior angles, and they are congruent. 2. Similarly, since AD is parallel to BC, angle A and angle B are also congruent. 3. Let's consider angle A. The bisector of angle A divides it into two congruent angles, let's call them angle AED and angle AEF. 4. Similarly, the bisector of angle B divides it into two congruent angles, let's call them angle BFD and angle BFE. 5. Since angle A and angle B are congruent, angle AED and angle BFD are also congruent, and angle AEF and angle BFE are congruent. 6. Now, let's consider triangle AED and triangle BFD. They have two pairs of congruent angles (angle AED = angle BFD and angle AEF = angle BFE) and a common side (AD). 7. By the Angle-Angle-Side (AAS) congruence criterion, triangle AED is congruent to triangle BFD. 8. Congruent triangles have congruent corresponding sides. Therefore, side DE is congruent to side DF and side AE is congruent to side BF. 9. Since DE is congruent to DF and AE is congruent to BF, triangle DEF is an isosceles triangle. 10. In an isosceles triangle, the altitude from the vertex angle bisects the base. Therefore, line segment EF is the altitude from the vertex angle D in triangle DEF, and it bisects side DE. 11. Since EF bisects side DE, angle EDF is a right angle (90°).

Therefore, we have proved that the bisectors of the angles in a parallelogram, which are adjacent to one side, intersect at a right angle (90°).

I hope this explanation helps! Let me know if you have any further questions.